Mean value theorem states that for continuous function f on [a, b] and differentiable on (a, b), there is at least one point $\displaystyle c \in (a, b)$ such that $\displaystyle f'(c) = \frac{f(b) - f(a)}{b-a}.$

Consider the function $\displaystyle f(x) = x^3.$

f'(x) = 0 at x = 0. However, there is no interval [a , b] that satisfies mean value theorem (since f(b) > f(a) for all b > a).

Why does the Mean value theorem fail here?